# Is $\phi B(\omega,\;\cdot\;)$ Lebesgue integrable over $[0,\infty)$ for a real-valued Brownian motion $B$ and $\phi\in C_c^\infty(\mathbb R)$?

Let $(B_t)_{t\ge 0}$ be a real-valued Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname P)$ and $\lambda$ be the Lebesgue measure on $\mathbb R$. Is $$W(\phi):=\int_{[0,\infty)}\phi(t)B_t\;{\rm d}\lambda\;\;\;\text{for }\phi\in C_c^\infty(\mathbb R)$$ well-defined?

Let $\phi\in C_c^\infty(\mathbb R)$ and $K:=\operatorname{supp}\phi\cap [0,\infty)$. I would argue that $$\left\|\phi1_{[0,\infty)}B(\omega,\;\cdot\;)\right\|_{L^1(\lambda)}\le\underbrace{\left\|\phi\right\|_{L^\infty(\lambda)}}_{<\infty}\underbrace{\left\|1_KB(\omega,\;\cdot\;)\right\|_{L^1(\lambda)}}_{=:M(\omega)}\;\;\;\text{for all }\omega\in\Omega$$ and that $M<\infty$ almost surely, since $K$ is compact and $B$ is almost surely continuous.

How do I deal with the null set on which $B$ is not continuous? Please note that $W$ is the distributional stochastic process associated with $B$. Maybe we don't care about its values on a null set.

$M<\infty$ because $B$ is almost surely continuous and thus bounded when restricted to a compact set
• That's what I've said, but we only have $M<\infty$ almost surely. – 0xbadf00d Jan 30 '16 at 12:39
• @0xbadf00d Almost surely is the most you can possibly say about any property of $B_t$. – David C. Ullrich Jan 30 '16 at 13:12
• @Bananach But it's essentially the same argument. Since $K$ is compact and $B$ is continuous, $B$ is Riemann integrable and the Riemann integral and the Lebesgue integral coincide. I've edited the question cause I want to make clear that my real question is how I need to handle the null set. – 0xbadf00d Jan 30 '16 at 23:57